The freeness theorem for equivariant cohomology of Rep( C 2 )-complexes
aa r X i v : . [ m a t h . A T ] J un THE FREENESS THEOREM FOR EQUIVARIANTCOHOMOLOGY OF REP( C )-COMPLEXES ERIC HOGLE AND CLOVER MAY
Abstract.
Let C be the cyclic group of order two. We show that the RO ( C )-graded Bredon cohomology of a finite Rep( C )-complex is free as amodule over the cohomology of a point when using coefficients in the constantMackey functor F . This paper corrects some errors in Kronholm’s proof ofthis freeness theorem. It also extends the freeness result to finite type com-plexes, those with finitely many cells of each fixed-set dimension. We give acounterexample showing the theorem does not hold for locally finite complexes. Contents
1. Introduction 12. Preliminaries 23. Computational tools 44. Change of basis for free modules 95. Change of basis for attaching a representation cell 146. Freeness theorem 167. Shifting of free generators 248. Kronholm’s proof 28References 321.
Introduction
Let C be the cyclic group of order two. We are concerned with Rep( C )-complexes, a class of C -spaces (see Definition 2.2). Our goal is to correct some sub-tle errors in Kronholm’s proof [6] that finite Rep( C )-complexes have free RO ( C )-graded Bredon cohomology in F -coefficients. We further demonstrate that Kro-nholm’s freeness theorem holds for finite type Rep( C )-complexes, those havingfinitely many cells of each fixed-set dimension (and hence also of each topologicaldimension). The freeness theorems lifts to a splitting at the spectrum level. Wegive a counterexample demonstrating that freeness does not necessarily hold forlocally finite Rep( C )-complexes.Kronholm’s freeness theorem is a powerful computational tool. It shows thecohomology of a finite Rep( C )-complex is free even in the presence of nonzero dif-ferentials corresponding to the attaching maps for representation cells. Kronholm’stheorem has been used by Dugger to study infinite C -equivariant Grassmanniansthat are finite type in [1] and by the first author to study finite C -equivariantGrassmannians in [5]. Prior to Kronholm’s work on the freeness theorem, Lewis [7] proved a freenesstheorem for the cohomology of Rep( C p )-complexes, where p is any prime. Lewisrequires the complexes have only even-dimensional cells with a further restrictionon the fixed-set dimensions in order to force all differentials in the cellular spec-tral sequence to be zero. Ferland [2], building on the work of Lewis, generalizedthe freeness theorem to finite type Rep( C p )-complexes with even-dimensional cellsfor p odd. Ferland’s result, like Kronholm’s, allows for nonzero differentials. Atodd primes, it is not possible to extend Ferland’s freeness theorem to include allfinite type Rep( C p )-complexes (see Counterexample 3.4). This makes it all themore surprising that Kronholm’s freeness theorem holds for all finite type Rep( C )-complexes.1.1. Organization.
In Section 2 we introduce the necessary background and no-tation, largely from [6]. In Section 3 we recall a number of computational toolsfrom [6] and [9]. The main proof will require a change of basis of a free module,the first step of which is given in Section 4. A further restriction on the changeof basis for the cohomology of a space is given in Section 5. These two steps aresimilar to Kronholm’s change of basis in [6]. In Section 6 we give a proof of Kro-nholm’s freeness theorem for finite complexes and extend the result to finite typecomplexes. We provide a counterexample to warn against overgeneralizing this re-sult. We also show the freeness theorems lift to a splitting at the spectrum level.Kronholm observed that a nontrivial differential causes generators to “shift” and inSection 7 we calculate these changes in the degrees of the generators of cohomologyafter a nontrivial differential. In Section 8 we explain in detail the main error inKronholm’s proof, which led to this paper. Much of our setup follows Kronholm’soriginal proof and the error there is quite subtle.1.2.
Acknowledgements.
The authors would like to thank Dan Dugger for in-troducing them to the beautiful subject of equivariant topology and for his supportthroughout many revisions. Thanks also to Mike Hill for many helpful conversa-tions, particularly regarding the finite type and locally finite cases. This work waspartially funded by the University of Oregon, UCLA, and Gonzaga University.2.
Preliminaries
To begin, we set up some basic machinery and notation much as in [6] with a fewsmall variations. Let G be a finite group. Given an orthogonal real G -representation V , let D ( V ) and S ( V ) denote the unit disk and unit sphere in V , respectively. Let S V = b V denote the representation sphere given by the one-point compactificationof V . There are two important types of equivariant cell complexes. Definition 2.1. A G -CW complex is a G -space X with a filtration, where X isa disjoint union of orbits G/H and X n is obtained from X n − by attaching cells ofthe form ( G/H α ) × D n along equivariant maps f α : G/H α × S n − → X n − . Thecells are attached via the usual pushout diagram ` α G/H α × S n − X n − ` α G/H α × D n X n ⊔ α f α where D n and S n − have the trivial G -action. REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 3 We will mainly be interested in another type of cell structure, one that is builtwith representation cells, called a Rep( G )-complex. Definition 2.2. A Rep( G ) -complex is a G -space X with a filtration X n where X is a disjoint union of trivial orbits of the form G/G = ∗ and X n is obtainedfrom X n − by attaching cells of the form D ( V α ) , where V α is an n -dimensional realrepresentation of G . The cells are attached along maps f α : S ( V α ) → X n − via theusual pushout diagram. The space X n in either filtration is referred to as the n -skeleton of X and thefiltration is referred to as a cell structure . If the filtration is finite, then X is finitedimensional . If there are finitely many cells of each dimension, then X is called locally finite . We call a Rep( C )-complex finite type if it has finitely many cellsof each fixed-set dimension, as defined below. If X is a connected Rep( G )-complex,the filtration quotients are wedges of representation spheres X n /X n − ∼ = W α S V α . Remark 2.3.
Any Rep( G )-complex can be given the structure of a G -CW complex.The converse is false. In particular, any G -space with a free action cannot be giventhe structure of a Rep( G )-complex. A Rep( G )-complex has at least one fixed pointbecause the origin of any real representation is fixed.We now specialize to the group G = C . As in [6], we write a p -dimensional real C -representation V as V ∼ = ( R , ) p − q ⊕ ( R , ) q = R p,q where R , is the trivial 1-dimensional real representation of C and R , is the signrepresentation. Allowing p and q to be integers if V is a virtual representation,we call p the topological dimension and q the weight or twisted dimension of V = R p,q . We will also refer to the fixed-set dimension , which is p − q . Wewrite S V = S p,q for the (possibly virtual) representation sphere given by theone-point compactification of V .For the V -th graded component of the ordinary RO ( C )-graded Bredon equivari-ant cohomology of a C -space X with coefficients in the constant Mackey functor F , we write H VG ( X ; F ) = H p,q ( X ; F ). We often suppress the coefficients andsimply write H p,q ( X ; F ) = H p,q ( X ). When we work nonequivariantly, H ∗ sing ( X )denotes the singular cohomology of the underlying topological space X with F -coefficients. The genuine equivariant Eilenberg–MacLane spectrum representing˜ H ∗ , ∗ ( − ) is H F . It has as its underlying spectrum H F . Given a homogeneouselement x ∈ H p,q ( X ), we use the notation | x | = ( p, q ) for the bidegree, top( x ) = p for the topological dimension, wt( x ) = q for the weight, and fix( x ) = p − q for thefixed-set dimension. It is often convenient to plot the bigraded cohomology in theplane. We will always plot the topological dimension p horizontally and the weight q vertically.With coefficients in the constant Mackey functor F , the cohomology of a pointwith the trivial C -action is the ring M := H ∗ , ∗ (pt) pictured in Figure 1. Onthe left is a more detailed depiction, however in practice it is easier to work with Note that the definition in [6] allows for X to be made up of any G -orbits. However, C isitself a C orbit and does not have free cohomology, which would contradict the freeness theorem. This is a departure from the usual notation. Kronholm [6], Shulman [11], and Ferland andLewis [3] use the notation | x | to denote the topological dimension p rather than the bidegree, andthe notation | x G | to denote the fixed-set dimension p − q . ERIC HOGLE AND CLOVER MAY the more succinct version on the right. Every lattice point inside the two “cones”represents a copy of F . There are unique nonzero elements ρ ∈ H , (pt) and τ ∈ H , (pt). Considered as an F [ ρ, τ ]-module, M splits as M = M +2 ⊕ M − where the top cone M +2 is a polynomial algebra with generators ρ and τ . There is aunique nonzero element in bidegree (0 , −
2) of the bottom cone M − . This element θ ∈ H , − (pt) is infinitely divisible by both ρ and τ and satisfies θ = 0. We saythat every element of the lower cone is both ρ -torsion and τ -torsion because it iskilled by a multiple of ρ and some multiple of τ . pq − − − − − − − ρτθ θρ θτ pq − − − − − − − M Figure 1. M = H ∗ , ∗ (pt; F ).The RO ( C )-graded cohomology H ∗ , ∗ ( X ) is a bigraded M -module. By M -module we always mean bigraded M -module, and any reference to an M -modulemap means a bigraded homomorphism. For a free M -module with a single gener-ator ω with bidegree | ω | = ( p, q ) we use the notation M h ω i = Σ p,q M . We write A n for the cohomology of S na , the n -dimensional sphere with the antipodal action,as an M -module. Notice that S na has a free C -action, so this is not an example ofRep( C )-complex. The M -module A n plays an important role in the cohomologyof C -CW complexes, but we will see that it cannot appear in the cohomology of aRep( C )-complex.A picture of A n (actually of A ) appears in Figure 2. Once again, on the left is amore detailed depiction, while in practice it is more convenient to use the succinctversion on the right. Here every lattice point in the infinite strip of width n + 1represents an F . Diagonal lines represent multiplication by ρ and vertical linesrepresent multiplication by τ , so that every nonzero element in A n is in the imageof τ and is not τ -torsion. We allow for n = 0 since C = S a has cohomology givenby a single vertical line. As a ring A n ∼ = F [ τ, τ − , ρ ] / ( ρ n +1 ), where multiplicationby ρ and τ corresponds to the module multiplication by the usual elements in M and where τ − has bidegree (0 , − Computational tools
In this section we present some tools for computing the RO ( C )-graded coho-mology of C -equivariant spaces. Let X be a connected Rep( C )-complex. Then REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 5 pq n ...... pq n ...... An Figure 2. A n = H ∗ , ∗ ( S na ; F ). X has a filtration coming from the cell structure where the filtration quotients X n /X n − are wedges of n -dimensional representation spheres corresponding to therepresentation cells that were attached.More generally, given any filtration of a C -space X pt = X ⊆ X ⊆ · · · ⊆ X k ⊆ X k +1 ⊆ · · · ⇒ X corresponding to the cofiber sequence X k ֒ → X k +1 → X k +1 /X k for each weight q there is a long exact sequence · · · → ˜ H p,q ( X k +1 /X k ) → ˜ H p,q ( X k +1 ) → ˜ H p,q ( X k ) d −→ ˜ H p +1 ,q ( X k +1 /X k ) → · · · . We often refer to the long exact sequences taken collectively for all q as “the longexact sequence.” Then d , taken collectively for all p and q , is a graded M -modulemap d : ˜ H ∗ , ∗ ( X k ) → ˜ H ∗ +1 , ∗ ( X k +1 /X k ), which we call the “differential” in the longexact sequence. This gives a short exact sequence of graded M -modules0 → cok( d ) → ˜ H ∗ , ∗ ( X k +1 ) → ker( d ) → . In many cases cok( d ) and ker( d ) are relatively easily determined, but computing˜ H ∗ , ∗ ( X k +1 ) requires solving the extension problem presented in this short exactsequence.As in the previous section, we plot RO ( C )-graded cohomology in the planewith the topological dimension p along the horizontal axis and the weight q alongthe vertical axis. The differential d in the long exact sequence is depicted by ahorizontal arrow since it increases topological dimension by one. When ˜ H ∗ , ∗ ( X k )is free as a graded M -module, i.e. when˜ H ∗ , ∗ ( X k ) ∼ = M h γ , . . . , γ k i = M i Σ | γ i | M , As Kronholm [6] points out, these long exact sequences sew together in the usual way to givea spectral sequence for each weight q . See Proposition 8.1 in Section 8. ERIC HOGLE AND CLOVER MAY the differential is determined by its image d ( γ i ) on the basis elements or on any setof generators.Before moving on we present an example that illustrates some common com-putational techniques as well as some advantages of the main theorem. We willreturn to this example in Section 8. In the computation presented here we use thefollowing fact from Section 6 in [9], which says we can compute the p -axis of the RO ( C )-graded cohomology of a space using singular cohomology of the quotient. Lemma 3.1.
Let X be a C -space. Then ˜ H p, ( X ) ∼ = H p sing ( X/C ) . Example 3.2.
In this example we compute the cohomology of the projective space R P tw = P ( R , ) using Lemma 3.1. A picture of R P tw is shown in Figure 3. This isthe usual depiction of a disk with opposite points on the boundary identified. The C -action is given by rotating the picture 180 ◦ .stuff Figure 3.
A depiction of R P tw .The long exact sequence associated to the cofiber sequence S , ֒ → R P tw → S , is depicted on the left side of Figure 4. Recall that in these depictions every latticepoint inside the cones represents an F .The map d is determined by its image on the generator of ˜ H ∗ , ∗ ( S , ) ∼ = Σ , M .It is necessarily nonzero because the quotient space R P tw /C is the cone on S ,which is contractible. The modules cok( d ) and ker( d ) resulting from this differen-tial are on the right side of Figure 4. Even knowing the differential, computing˜ H ∗ , ∗ ( R P tw ) requires solving the extension problem in the short exact sequence0 → cok( d ) → ˜ H ∗ , ∗ ( R P tw ) → ker( d ) → . It is not at all obvious at this stage that the solution to this extension prob-lem should be a free M -module. However, since R P tw is a Rep( C )-complex˜ H ∗ , ∗ ( R P tw ) must be free as a result of the freeness theorem in Section 6.1. Thusthe cohomology of R P tw as an M -module is ˜ H ∗ , ∗ ( R P tw ) = Σ , M ⊕ Σ , M , aspictured in Figure 5.Notice that in the cohomology of R P tw there are two copies of M generated inthe same topological dimensions as before, but they have “shifted.” One generatornow has higher weight and the other lower weight. This is an example of a moregeneral behavior known as a Kronholm shift, described in Section 7. We giveformulas that precisely quantify these shifts in Theorem 7.1. The power of thefreeness theorem and shifting formulas is that when a nonzero differential like theone above occurs, the resulting cohomology must be free with generators appearingto have shifted weights appropriately. Aside 3.3.
One might expect a similar freeness result to hold more generally forRep( C p )-complexes. However, for odd primes one quickly discovers a space analo-gous to R P tw that does not have free cohomology. This makes it rather surprising REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 7 pq − − − − − − − d pq − − − − − − − ker( d )cok( d ) RP2
Figure 4.
Differential in a long exact sequence for ˜ H ∗ , ∗ ( R P tw ). pq − − − − − − − RP2
Figure 5.
Reduced cohomology of R P tw .that the freeness theorem holds for all finite Rep( C )-complexes. For concreteness,we take p = 3 and give an example below of a Rep( C )-complex whose cohomologyis not free. Counterexample 3.4.
Let X be the C -space whose underlying space is a 2-simplex with edges identified as pictured in Figure 6. Here a generator of C actsby rotating the picture 120 ◦ .The space X can be realized as a Rep( C )-complex. Using techniques similarto those in Example 3.2, one can compute the cohomology of X with constant F -coefficients as a module over the cohomology of a point. Of course, the cohomologyof a point as a C -space with F coefficients is not M , but it shares several prop-erties with M and can also be depicted with two cones. One can readily verify thecohomology of X is not free as a module over this ring.We now return to the prime two and assemble a few more computational tools.We will use several results from [9] to simplify the proof of the freeness theorem. In ERIC HOGLE AND CLOVER MAY stuff
Figure 6. A C analogue of R P tw .particular, from [9] we have the following structure theorem for the RO ( C )-gradedcohomology of C -CW complexes. The structure theorem says that as a moduleover the cohomology of the point, the RO ( C )-graded cohomology of a finite C -CW complex decomposes as a direct sum of two basic pieces: shifted copies of thecohomology of a point and shifted copies of the cohomologies of spheres with theantipodal action. Theorem 3.5 (Structure Theorem) . For any finite C -CW complex X , there is adecomposition of the RO ( C ) -graded cohomology of X as an M -module given by H ∗ , ∗ ( X ; F ) ∼ = M i Σ p i ,q i M ! ⊕ M j Σ r j , A n j , where ≤ q i ≤ p i and ≤ r j , n j . The goal of this paper is to show that for the special case of a finite Rep( C )-complex, the cohomology is free. That is, we will show the cohomology containsonly shifted copies of M and not any copies of A n .From the structure theorem we immediately obtain a description of the localiza-tions of the cohomology of a finite C -CW complex. Notice that A n is preserved by τ -localization while τ − M ∼ = A ∞ ∼ = F [ τ ± , ρ ]. On the other hand ρ -localizationkills A n and ρ − M ∼ = F [ τ, ρ ± ]. Corollary 3.6.
Let X be a finite C -CW complex with H ∗ , ∗ ( X ) ∼ = M i Σ p i ,q i M ! ⊕ M j Σ r j , A n j . Then τ − H ∗ , ∗ ( X ) ∼ = M i Σ p i , A ∞ ! ⊕ M j Σ r j , A n j and ρ − H ∗ , ∗ ( X ) ∼ = M i Σ p i − q i , (cid:0) ρ − M (cid:1) . Remark 3.7.
Thus for X a finite complex, H ∗ , ∗ ( X ) is free if and only if τ − H ∗ , ∗ ( X )has no ρ -torsion. Furthermore, ρ − H ∗ , ∗ ( X ) depends only on the fixed-set dimen-sions of the free generators in H ∗ , ∗ ( X ). REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 9 In addition to knowing the cohomology is free, we would like to know where thefree generators live. The cohomology of any finite C -CW complex has “vanishingregions” where the cohomology is zero. The following proposition and corollaryappear in [9]. Proposition 3.8.
Let X be a finite C -CW complex of dimension m . Then H p,q ( X ) = 0 whenever (1) p < and q > p − , or (2) p > m and q < p − m . In particular, we have the immediate corollary.
Corollary 3.9.
Any generator of a copy of M in the cohomology of a finite m -dimensional C -CW complex X must lie in a bidegree ( p, q ) corresponding to anactual representation with topological dimension p satisfying ≤ p ≤ m and weight ≤ q ≤ p . The region where M generators can lie is depicted by the triangle in Figure 7on the right. To find M generators we may use θ as in the following lemma from[9]. − pq m m Figure 7.
Vanishing regions and region containing M generators. Lemma 3.10.
If a graded M -module contains a nonzero homogenous element x with θx nonzero, then M h x i is a graded free submodule. In [9] it is also shown that M is self-injective. Lemma 3.11.
The regular module M is injective as a graded M -module. Thus to find free summands of an M -module, it is often useful to find an elementwith a nontrivial θ -multiple. Such an element generates a free submodule, whichsplits off as a direct summand because M is self-injective.4. Change of basis for free modules
Following in Kronholm’s footsteps, in the proof of the freeness theorem in Section6 we will induct on the number of cells of a Rep( C )-complex and attach one cellat a time. For the inductive step, we will need to consider differentials from a freemodule to a single shifted M corresponding to the newly attached cell. Kronholm’s proof includes a change of basis lemma that simplifies the differentials in this setting[6, Lemma 3.1]. However, there is a small error in the proof of this lemma. Forcompleteness, we will first prove an algebraic version of the change of basis inspiredby Kronholm’s argument. In Section 5, we use the algebraic version and a factabout ρ -localization to easily deduce the change of basis lemma that appears in [6].We now show that given a nonzero homomorphism from a graded free module Γto Σ p,q M , a basis for Γ can be chosen in one of two particularly nice ways: eitherthere is a single basis element with a nonzero map into the top cone of the target M or there is a “ramp” of nonzero maps into the bottom cone, as described in [6].These possibilities are listed in the following lemma. Lemma 4.1. (Change of basis) Consider a graded M -module homomorphism withbidegree (1 , of the form Γ = M h γ , γ , . . . , γ m i d −→ M h ν i where | ν | = ( p, q ) .Assume for all i that top( γ i ) ≤ p and whenever top( γ i ) = p then wt( γ i ) ≤ q . Thenthere is a change of basis for Γ so that either (1) d ≡ (and no change of basis is required); (2) Γ ∼ = M h λ, χ , . . . , χ m i with • d ( χ i ) = 0 for all i , • d ( λ ) is nonzero, and • top( λ ) = p − and wt( λ ) ≥ q ;so that λ is the only basis element supporting a nonzero image; or (3) Γ ∼ = M h ω , . . . , ω n , χ n +1 , . . . , χ m +1 i with • d ( χ i ) = 0 for all i , • d ( ω i ) is nonzero for all i , • wt( ω i ) ≤ q − , • top( ω i ) < top( ω i +1 ) , and • fix( ω i ) < fix( ω i +1 ) ;so that ω , . . . , ω n are the only basis elements with nonzero images and theyhave increasing topological dimension and increasing fixed-set dimension. In the second case, d ( λ ) lands in the upper cone M +2 h ν i , while in the third case ω , . . . , ω n map to nonzero elements of M − h ν i . Examples of these two possibilitiesare depicted in Figure 8 (with the χ i omitted). In these pictures, the topologicaldimension of a basis element determines the horizontal position and the weightdetermines the height. The fixed-set dimension determines the line of slope one onwhich the generator lies and d , which maps red to blue, is determined by its imageon basis elements.The positions of the basis elements ω , . . . , ω n are referred to as a ramp oflength n in [6] (or “stairstep” pattern in [3]). We say that ω , . . . , ω n satisfy the ramp condition , i.e. have strictly increasing topological dimension and fixed-setdimension. Thus each ω i +1 is to the right of ω i and on a lower diagonal. The rampcondition also implies that no ω i lies in a bidegree that sits inside the upper cone M +2 h ω j i for i = j . An example of M generators satisfying the ramp condition isdepicted in Figure 9. Proof.
For d ≡ d is nontrivial.Partition a basis for Γ, reordering if necessary, as { γ , . . . , γ r } ∪ { γ r +1 , . . . , γ r + s } ∪ { γ r + s +1 , . . . , γ r + s + t } A more significant error in Kronholm’s main proof will be explained in Section 8.
REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 11 pq d λ ν pqq − d d d ω ω ω ν Figure 8.
Examples of cases (2) and (3) in Lemma 4.1. ω ω ω ω ω Figure 9.
A ramp of length 5.so that the basis elements γ , . . . , γ r have nonzero images in the top cone M +2 h ν i ,basis elements γ r +1 , . . . , γ r + s have nonzero images in the bottom cone M − h ν i , and d is zero on γ r + s +1 , . . . , γ r + s + t . We allow for the possibility that either r = − s = 0 if there are nonein the bottom cone, or t = 0 if there are no generators mapped to zero. Set∆ + = M h γ , . . . , γ r i ∆ − = M h γ r +1 , . . . , γ r + s i ∆ = M h γ r + s +1 , . . . , γ r + s + t i so that Γ ∼ = ∆ + ⊕ ∆ − ⊕ ∆ .We first assume there are indeed nonzero images in the top cone. We will showthere is a change of basis for ∆ + = M h γ , . . . , γ r i so that λ, χ , . . . , χ r is a basissatisfying d ( λ ) = 0 but d ( χ i ) = 0 for each i . For degree reasons, any γ i withnonzero image in the top cone of ν has top( γ i ) = p −
1. So for 0 ≤ i ≤ r , the image d ( γ i ) = τ k i ν for k i ≥
0. Reindexing if necessary, we can assume γ is of minimalweight among γ , . . . , γ r . Rename it λ := γ so that d ( λ ) = τ k ν and k i ≥ k for all i . Since λ has the lowest weight, each remaining γ i lies in the upper cone of λ as in Figure 10. pq νγ i λ Figure 10.
Nonzero images in the upper cone.We can use λ to form the χ i . For each i with 1 ≤ i ≤ r , replace γ i with χ i := γ i + τ k i − k λ . Notice that χ i is a homogeneous element in the same bidegreeas γ i and that χ i ∈ ker( d ). The χ i are all independent, as a dependence amongthem would give rise to a dependence among the γ i . Indeed, suppose there existednonzero elements M i ∈ M for 1 ≤ i ≤ r such that each M i χ i is in the samebidegree and P M i χ i = 0. Then r X i =1 M i γ i + r X i =1 M i τ k i − k λ = r X i =1 M i γ i + r X i =1 M i τ k i − k ! γ = 0 . The coefficient of γ could be zero, but there is at least one γ i with i > M i . Thus we have a dependence among the basis elements γ i ,a contradiction. We can therefore include a free module and quotient to get theshort exact sequence 0 → r M i =1 M h χ i i → ∆ + → Q → , where Q ∼ = M h λ i . So ∆ + is isomorphic to M h λ, χ , . . . , χ r i since M is self-injective. Now we have a basis for ∆ + with d ( λ ) = 0 but d ( χ i ) = 0 for each i . We use a similar approach to make a change of basis for ∆ − = M h γ r +1 , . . . , γ r + s i ,now using λ to modify the basis elements with nonzero images in the bottom cone.For r + 1 ≤ i ≤ r + s , we have d ( γ i ) = θρ ji τ ki ν for some j i , k i ≥ γ i ) ≤ p − γ i ) ≤ q −
2. In particular, for r + 1 ≤ i ≤ r + s ,each basis element γ i has a bidegree that lies inside the lower cone of λ as in Figure11. Not every combination of the γ i is independent. For example, { γ , γ + θγ } is dependentbecause τγ = τ ( γ + θγ ). REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 13 pqq − νγ i λ Figure 11.
Images from ∆ − and ∆ + .Recall that we defined λ := γ above so d ( λ ) = τ k ν . Replace the basis element γ i (for r + 1 ≤ i ≤ r + s ) with χ i := γ i + θρ ji τ ki + k λ . Again the coefficient of λ has been chosen so that χ i is a homogenous element generating an M in the samebidegree as γ i and d ( χ i ) = 0. As before, the χ i are independent. Thus we have anew basis for ∆ − given by χ r +1 , . . . , χ r + s .Finally, for ∆ = M h γ r + s +1 , . . . , γ r + s + t i no real change of basis is required.The map d is already zero on each basis element, so simply define χ i := γ i for r + s + 1 ≤ i ≤ r + s + t . Taking the union of the new bases, we now have a basisfor Γ ∼ = ∆ + ⊕ ∆ − ⊕ ∆ of the form λ, χ , . . . , χ m satisfying the conditions in case(2) of the lemma.It remains to deal with case (3). The change of basis above depended on havingsome nonzero image in the top cone. If instead we have that r = − + is zero,then Γ ∼ = ∆ − ⊕ ∆ , where ∆ − = M h γ , . . . , γ s − i and ∆ = M h γ s , . . . , γ s + t − i .In this case, we will show that we can find a ramp of basis elements ω , . . . , ω n supporting nonzero images in the bottom cone as described in case (3).Consider any two distinct basis elements γ a and γ b , 0 ≤ a, b ≤ s −
1. Since γ a and γ b support nonzero images in the bottom cone of ν , we can write d ( γ a ) = θρ ja τ ka ν and d ( γ b ) = θρ jb τ kb ν . If γ a and γ b do not satisfy the ramp condition thenone lies inside the range of the upper cone of the other, as shown in Figure 12.Without loss of generality, we can assume γ b lies inside the upper cone of γ a so thattop( γ a ) ≤ top( γ b ) and fix( γ a ) ≥ fix( γ b ).Multiplication by ρ preserves fixed-set dimension but increases topological di-mension by one. Multiplication by τ preserves topological dimension but decreasesfixed-set dimension by one. Thus j a ≥ j b and k a ≥ k b . In the basis for ∆ − replace γ b with χ b := ρ j a − j b τ k a − k b γ a + γ b . Once more χ b is defined to be homogeneous inthe same bidegree as γ b and d ( χ b ) = 0. As before, the independence of γ a and γ b will imply that γ a and χ b are independent as well. So M h γ a , γ b i ∼ = M h γ a , χ b i .Continue to reduce the set of γ i supporting nonzero images in the bottom cone inthis way until no γ b lies in the upper cone region of any γ a . The remaining γ i withnonzero images satisfy the ramp condition up to reindexing. Reindex the remaining pqq − νγ a γ b Figure 12.
Differentials to the lower cone. γ i to have increasing topological dimension, rename ω i := γ i and reindex the χ i asnecessary. We now have a basis for ∆ − of the form ω , . . . , ω n , χ n +1 , . . . , χ s − .Finally for ∆ = M h γ s , . . . , γ s + t − i , as before, no real change of basis is requiredas d is already zero on each basis element. Again define χ i := γ i for s ≤ i ≤ s + t − ∼ = ∆ − ⊕ ∆ of the form ω , . . . , ω n , χ n +1 , . . . , χ m +1 satisfying the conditions in case (3) of the lemma. (cid:3) The details of the ramp condition in the previous lemma will not play a big rolein the proof of the freeness theorem, however they will be crucial in Section 7 wherewe determine the bidegrees of generators after a nontrivial differential.5.
Change of basis for attaching a representation cell
In the previous section, we introduced an algebraic change of basis motivatedby Kronholm’s argument. In this section we will deduce some consequences forthe cohomology of a Rep( C )-complex as an M -module, completing the proof ofLemma 3.1 from [6].First we recall a lemma and remark from [9] that localization by ρ relates equi-variant cohomology to the singular cohomology of the fixed set, which will restrictthe types of differentials we may encounter. Lemma 5.1. ( ρ -localization) Let X be a finite C -CW complex. Then ρ − H ∗ , ∗ ( X ) ∼ = ρ − H ∗ , ∗ ( X C ) ∼ = H ∗ sing ( X C ) ⊗ F ρ − M . Remark 5.2.
An important consequence of the previous lemma is that ρ − ˜ H ∗ , ∗ ( X )does not have any τ -torsion, since ρ − M ∼ = F [ τ, ρ ± ] and ρ − ˜ H ∗ , ∗ ( X ) is free as a ρ − M -module.We will use this fact to show the long exact sequence for attaching a singlecell to a Rep( C )-complex cannot have any non-surjective differentials into the topcone. That is, if there is a nontrivial differential into the top cone as in the secondcase of Lemma 4.1 so that d ( λ ) = τ k ν for some k , then in fact wt( λ ) = wt( ν ), k = 0, and the restriction M h λ i → M h ν i is an isomorphism. The following REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 15 lemma looks similar to Lemma 4.1, but the content here is a further restriction ondifferentials into the top cone for the cohomology of Rep( C )-complexes, rather than M -modules in general. The result is quite similar to Lemma 3.1 in [6] althoughthe proof differs slightly. Lemma 5.3.
Let B be a Rep( C ) -complex with reduced cohomology given by agraded free M -module with basis γ , . . . , γ m so ˜ H ∗ , ∗ ( B ) ∼ = M h γ , . . . , γ m i . Suppose X is obtained from B by attaching a single ( p, q ) cell and let ν denote the generatorfor the reduced cohomology of X/B ∼ = S p,q . Assume further that for all i , top( γ i ) ≤ p and whenever top( γ i ) = p then wt( γ i ) ≤ q . The cofiber sequence B ι −→ X π −→ S p,q gives rise to the differential d : ˜ H ∗ , ∗ ( B ) → ˜ H ∗ +1 , ∗ ( S p,q ) in the long exact sequence.After an appropriate change of basis, one of the following is true: ( a ) the differential d ≡ ; ( b ) the differential is zero on every basis element except λ , where | λ | = ( p − , q ) and the restriction of the differential to M h λ i → M h ν i is an isomorphism;or ( c ) the basis elements supporting nonzero differentials satisfy the ramp condi-tion and map to M − h ν i .Proof. Change the basis for ˜ H ∗ , ∗ ( B ) according to Lemma 4.1. Recall from the proofof the lemma that the change of basis preserved the bidgrees of the generators. Soany new basis element x still satisfies top( x ) ≤ p and if top( x ) = p then wt( x ) ≤ q .Case ( a ) corresponds to part (1) of Lemma 4.1. Now we will show case ( b ) followsfrom case (2). In this situation, we may assume there is a nonzero differentialinto the top cone M +2 h ν i . Recall that Lemma 4.1 says we can rewrite ˜ H ∗ , ∗ ( B ) ∼ = M h λ, χ , . . . , χ m i where top( λ ) = p −
1, wt( λ ) ≥ q , and d ( χ i ) = 0 for all i . Ifwt( λ ) = q , then the restriction of d to M h λ i → M h ν i is an isomorphism. Itremains to show this is the only possibility. Suppose instead that wt( λ ) > q so that d ( λ ) = τ k ν for some k >
0. Then the following equation holds. d (cid:18) θτ k λ (cid:19) = θτ k d ( λ ) = θτ k τ k ν = θν A portion of cok( d ) and ker( d ) are depicted in Figure 13. (All of the χ i are also inker( d ) but are not depicted here.)To actually compute ˜ H ∗ , ∗ ( X ) would require solving an extension problem ofgraded M -modules as ˜ H ∗ , ∗ ( X ) is in the middle of the short exact sequence0 → cok( d ) → ˜ H ∗ , ∗ ( X ) → ker( d ) → . However, we will not need to solve this extension problem because the ρ -localizationof ˜ H ∗ , ∗ ( X ) is actually independent of the resolution. By Remark 5.2, we know ρ − ˜ H ∗ , ∗ ( X ) cannot have any τ -torsion. Since τ k − ν / ∈ im( d ), by exactness wehave τ k − ν / ∈ ker( π ∗ ) and hence a nonzero class π ∗ ( τ k − ν ) ∈ ˜ H ∗ , ∗ ( X ). This classhas no ρ -torsion and thus survives ρ -localization. Yet it does have τ -torsion since Here we deviate from Kronholm, as there is a small error in the calculation of cok( d ) andker( d ) in [6]. Without the assumption that for all i top( γ i ) ≤ p and if top( γ i ) = p then wt( γ i ) ≤ q , it ispossible to have differentials that introduces ρ -torsion in ˜ H ∗ , ∗ ( X ). See Example 6.6 in [9]. pq d λ ν pq ker( d ) cok( d ) Figure 13.
Differential to the upper cone. τ · π ∗ ( τ k − ν ) = π ∗ ( τ k ν ) = 0, contradicting Lemma 5.1. Therefore case (2) ofLemma 4.1 reduces to case ( b ).Finally, if ˜ H ∗ , ∗ ( B ) satisfies the ramp condition of case (3) in Lemma 4.1, thencase ( c ) follows immediately. (cid:3) Freeness theorem
We are now ready to prove Kronholm’s freeness theorem for finite
Rep ( C )-complexes. The proof will proceed by induction on the number of representationcells and consider the attaching map for a single cell. The bulk of the proof willinvolve case ( c ) of Lemma 5.3, with a ramp of M generators supporting differentialsto the lower cone of another generator. This will lead to the cohomology being afree module with the same number of M generators, but in shifted bidgrees fromtheir original positions.In this main case, we use τ -localization to show that the cohomology is free.However, we will need to show the inductive hypothesis holds, namely that anygenerators in the highest topological dimension are below a certain weight. Forthat, we use τ -localization together with ρ -localization. Despite the generatorsappearing in new bidegrees, the set of topological dimensions of the generators aswell as the set of fixed-set dimensions are preserved. This will give us the constrainton the weights required to complete the inductive step.Somewhat surprisingly, in this main case we prove the cohomology is free withoutever identifying the free generators or the shifts. For the purpose of computations,a choice of free basis and a precise formula for calculating the Kronholm shifts aregiven in Section 7. Theorem 6.1. (Freeness theorem) If X is a finite Rep( C ) -complex then ˜ H ∗ , ∗ ( X ; F ) is free as a graded M -module, where M = H ∗ , ∗ (pt; F ) .Proof. Attaching one cell at a time, we can filter X so that each X k +1 is formedfrom X k by attaching a single Rep( C )-cell, e k +1 . Order the representation cells e , e , . . . , e K by increasing topological dimension and increasing weight so that if e i ∼ = D ( R p i ,q i ) then p i ≤ p i +1 for all i , and if p i = p i +1 then q i ≤ q i +1 . We proceed REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 17 by induction on the spaces in the ‘one-at-a-time’ cellular filtration pt = X ⊆ X ⊆ · · · ⊆ X k ⊆ X k +1 ⊆ · · · ⊆ X K = X. The base case is trivial.We will inductively prove that each ˜ H ∗ , ∗ ( X k ) is a free M -module, where all thegenerators have topological dimensions between zero and p k and any generators intopological dimension p k have weights between zero and q k . For the inductive stepwe will form X k +1 from X k by attaching a single ( p k +1 , q k +1 )-cell. In the courseof the inductive step we will see that the generator of ˜ H ∗ , ∗ ( X k +1 ) correspondingto this new cell may remain in bidegree ( p k +1 , q k +1 ), “shift” to a lower weight, orvanish.Now assume the inductive hypothesis holds for ˜ H ∗ , ∗ ( X k ). To simplify notationwe set p = p k +1 and q = q k +1 . We will show ˜ H ∗ , ∗ ( X k +1 ) is free with all generatorsin topological dimension p having weight at most q .The cofiber sequence X k ι −→ X k +1 π −→ X k +1 /X k ∼ = S p,q induces the long exact sequence · · · d −→ ˜ H ∗ , ∗ ( S p,q ) π ∗ −→ ˜ H ∗ , ∗ ( X k +1 ) ι ∗ −→ ˜ H ∗ , ∗ ( X k ) d −→ ˜ H ∗ +1 , ∗ ( S p,q ) π ∗ −→ · · · . Let ν denote the free generator of ˜ H ∗ , ∗ ( X k +1 /X k ) ∼ = ˜ H ∗ , ∗ ( S p,q ) ∼ = Σ p,q M inbidegree ( p, q ). To show that ˜ H ∗ , ∗ ( X k +1 ) is a free M -module we will solve theextension problem of graded M -modules that appears in the short exact sequence0 → cok( d ) → ˜ H ∗ , ∗ ( X k +1 ) → ker( d ) → . We proceed by investigating the differential d : ˜ H ∗ , ∗ ( X k ) → ˜ H ∗ +1 , ∗ ( S p,q ).Lemma 5.3 enumerates the possibilities. In case (a), the differential d ≡ → M h ν i → ˜ H ∗ , ∗ ( X k +1 ) → ˜ H ∗ , ∗ ( X k ) → , which splits since ˜ H ∗ , ∗ ( X k ) is free by inductive assumption. Hence ˜ H ∗ , ∗ ( X k +1 ) ∼ =˜ H ∗ , ∗ ( X k ) ⊕ M h ν i is free and there are no shifts. That is, the generators of˜ H ∗ , ∗ ( X k +1 ) are in the same bidegrees as the generators of ˜ H ∗ , ∗ ( X k ), with a singlenew generator in bidegree ( p, q ). In particular, since the inductive hypothesis holdsfor the generators of ˜ H ∗ , ∗ ( X k ), any free generators of ˜ H ∗ , ∗ ( X k +1 ) with topologi-cal dimension p will have weight not exceeding q . Thus ˜ H ∗ , ∗ ( X k +1 ) satisfies theinductive hypothesis.In case ( b ), the extension problem is again easily solved. Recall that after thechange of basis, ˜ H ∗ , ∗ ( X k ) ∼ = M h λ, χ , . . . , χ m i with | λ | = ( p − , q ) and d ( χ i ) = 0for all i . As the differential is surjective, cok( d ) = 0 and the short exact sequencebecomes 0 → ˜ H ∗ , ∗ ( X k +1 ) → M h χ , . . . , χ m i → , so ˜ H ∗ , ∗ ( X k +1 ) ∼ = M h χ , . . . , χ m i is free. Here we say λ kills ν , making ˜ H ∗ , ∗ ( X k +1 )the same as ˜ H ∗ , ∗ ( X k ), but with a single copy of M generated in bidegree ( p − , q ) Although we filter X using the ‘one-at-a-time’ cellular filtration as in Kronholm’s proof [6],we prove the theorem using the long exact sequence associated to a cofiber sequence with noreference to the cellular spectral sequence. This is addressed in more detail in Section 8. removed. Again, any free generators of ˜ H ∗ , ∗ ( X k +1 ) with topological dimension p have weight not exceeding q , satisfying the inductive hypothesis.It remains to solve the extension problem arising in case ( c ), which is significantlymore labor-intensive. Under these circumstances the usual short exact sequence0 → cok( d ) → ˜ H ∗ , ∗ ( X k +1 ) → ker( d ) → H ∗ , ∗ ( X k +1 ) is a free M -module. In Section7 we will see that some of the generators have shifted, i.e. given rise to generatorsin different bidegrees. However, it will still be the case that all free generatorshave topological dimensions between zero and p , and that any free generators withtopological dimension p have weight not exceeding q .In this case we can write ˜ H ∗ , ∗ ( X k ) ∼ = M h ω , . . . , ω n , χ n +1 , . . . , χ m +1 i where ω , . . . , ω n satisfy the ramp condition, each ω i supports a nonzero differential tothe bottom cone M − h ν i , and d ( χ i ) = 0 for all n + 1 ≤ i ≤ m + 1. The image under d of each ω i is d ( ω i ) = θρ j i τ k i ν for some integers j i , k i ≥
0. It follows from the ramp condition that j i > j i +1 and k i < k i +1 . That is, each ω i +1 is to the right of ω i and lies on a lower diagonal. Anexample of such a differential (with the χ i omitted) is pictured in Figure 14. pqq − νω ω ω ω ω Figure 14.
Ramp of differentials ( χ i omitted and n = 5).We will use τ -localization to show the solution to the extension problem in thiscase must be free. Since localization is exact, we may consider the long exactsequence · · · → τ − ˜ H ∗ , ∗ ( X k +1 ) → τ − ˜ H ∗ , ∗ ( X k ) τ − d −−−→ τ − ˜ H ∗ +1 , ∗ ( S p,q ) → · · · . Recall the ramp condition means the ω i have increasing topological dimension and increasingfixed-set dimension so top( ω i ) < top( ω i +1 ) and fix( ω i ) < fix( ω i +1 ). REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 19 Notice that im( d ) ⊆ M − h ν i so τ − d ≡
0. Thus this long exact sequence reduces toa short exact sequence0 → τ − ˜ H ∗ , ∗ ( S p,q ) → τ − ˜ H ∗ , ∗ ( X k +1 ) → τ − ˜ H ∗ , ∗ ( X k ) → , which we can rewrite as0 → Σ p,q A ∞ → τ − ˜ H ∗ , ∗ ( X k +1 ) → M i Σ | ω i | A ∞ ! ⊕ M j Σ | χ j | A ∞ → . As modules over τ − M ∼ = A ∞ , both the left and right terms are free. So this shortexact sequence splits and the middle term is also free. Notice X k +1 is also a finite C -CW complex so the structure theorem applies. By Remark 3.7, ˜ H ∗ , ∗ ( X k +1 ) hasno copies of A n for any n , making it a free M -module.At this point it may seem that we are done, but in fact we still need to verify theinductive hypothesis about the topological dimensions of the generators and thatthe weight of any generator in dimension p is no more than q . The splitting of theshort exact sequence above implies that all the free generators of ˜ H ∗ , ∗ ( X k +1 ) are inthe same topological dimensions as the free generators of ˜ H ∗ , ∗ ( S p,q ) and ˜ H ∗ , ∗ ( X k ).This might seem to suggest the generators have not shifted at all, but a shift up ordown would not be witnessed by τ -localization. In Section 7, we will see they haveindeed shifted.Knowing the topological dimensions of the generators, we use ρ -localization toprove the constraint on the weight of any generators in dimension p . First, we splitoff the less interesting free generators corresponding to the χ i ∈ ˜ H ∗ , ∗ ( X k ). The χ i are elements of ker( d ) so we can lift them back to ˜ H ∗ , ∗ ( X k +1 ). Since d ( χ i ) = 0,by exactness we can choose b i in the preimage ( ι ∗ ) − ( χ i ), though this choice isnot necessarily unique. It is fairly easy to see that θb i = 0 for all i . This followsbecause M h χ n +1 , . . . , χ m +1 i is a free submodule of ˜ H ∗ , ∗ ( X k ) so θχ i = 0 for each i . Since ι ∗ is an M -module map, ι ∗ ( θb i ) = θι ∗ ( b i ) = θχ i = 0. So it must bethat θb i = 0. Then by Lemma 3.10 each M h b i i includes as a free submodule of˜ H ∗ , ∗ ( X k +1 ). Using the independence of the χ i , we can take all the b i together andinclude M h b n +1 , . . . , b m +1 i as a free submodule.We now have a short exact sequence0 → M h b n +1 , . . . , b m +1 i → ˜ H ∗ , ∗ ( X k +1 ) → N → , where N is the quotient. By Lemma 3.11, we know that M is self-injective. So thisshort exact sequence splits, making N a direct summand of ˜ H ∗ , ∗ ( X k +1 ). Moreover,since we already know ˜ H ∗ , ∗ ( X k +1 ) is free, N is also free. Now we will ρ -localize the original long exact sequence. As with τ -localization, ρ -localization is exact and since im( d ) ⊆ M − h ν i we have that ρ − d ≡
0. So wehave the short exact sequence0 → ρ − ˜ H ∗ , ∗ ( S p,q ) → ρ − ˜ H ∗ , ∗ ( X k +1 ) → ρ − ˜ H ∗ , ∗ ( X k ) → . Using the ρ -localization of the short exact sequence involving N , we obtain a shortexact sequence of short exact sequences as in the diagram below. Everything in thefollowing diagram is a free ρ − M -module so all the short exact sequences split. Note that M is a graded local ring with unique maximal ideal ( ρ, τ ). Thus every finitelygenerated projective M -module is free. See [8]. L i ( ρ − M h b i i ) L i ( ρ − M h χ i i ) 00 ρ − ˜ H ∗ , ∗ ( S p,q ) ρ − ˜ H ∗ , ∗ ( X k +1 ) ρ − ˜ H ∗ , ∗ ( X k ) 00 ρ − ˜ H ∗ , ∗ ( S p,q ) ρ − N L i ( ρ − M h ω i i ) 00 0 0In particular, from the last row we have that ρ − N ∼ = (cid:16) ρ − ˜ H ∗ , ∗ ( S p,q ) (cid:17) ⊕ n M i =1 ( ρ − M h ω i i ) ! . Thus, making use of Remark 3.7, the fixed-set dimensions of the free generators of N are the same as the fixed-set dimensions of the free generators of M h ω , . . . , ω n i together with that of ˜ H ∗ , ∗ ( S p,q ).Again, it seems plausible at this point that we simply have a new generator inthe bidegree of ν with no shifts. However, as previously mentioned, we will seein Section 7 the generators have shifted. Regardless, we claim any free generatorof N in topological dimension p must have fixed-set dimension at least p − q andso have weight at most q (but possibly lower). This is because each of the ω i maps to the lower cone of ν , so fix( ν ) = p − q is the lowest fixed-set dimension ofthe generators of N . The generators b i are in precisely the same bidegrees as theoriginal generators χ i . Thus all free generators of ˜ H ∗ , ∗ ( X k +1 ) having topologicaldimension p have weight at most q . This completes the inductive step and hencethe proof of the freeness theorem for finite Rep( C )-complexes. (cid:3) Finite type freeness theorem.
As a consequence of the finite case, we canextend Kronholm’s freeness theorem to infinite complexes of finite type. Recall, wesay that a Rep( C )-complex is finite type if it is built with finitely many cells ofeach fixed-set dimension. Theorem 6.3. If X is a finite type Rep( C ) -complex then ˜ H ∗ , ∗ ( X ; F ) is free asa graded M -module.Proof. As in the proof of Theorem 6.1, we begin by filtering X as a sequenceof subcomplexes X k where each X k +1 is formed from X k by attaching a singleRep( C )-cell e k +1 . We assume the cells are attached by increasing topologicaldimension and, within each topological dimension, by increasing weight. We willshow the lim ←− term vanishes and find a free basis for the inverse limit.Because X is finite type, for each i ≥ X k ( i ) thatcontains every cell of fixed-set dimension less than or equal to i by defining k ( i ) = max { k : e k ∼ = D ( R p k ,q k ) and p k − q k ≤ i } . REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 21 Note that ˜ H ∗ , ∗ ( X k ( i ) ) is free by Theorem 6.1 because X k ( i ) is a finite Rep( C )-complex. Recall from the proof of Theorem 6.1 that the M generators of thecohomology may lie in different bidegrees than the cells forming X k ( i ) and there maybe fewer M generators than cells. However, each M generator of the cohomologywill have the same fixed-set dimension as one of the cells used in constructing thespace.Now we will consider the cofiber sequence X k ( i ) → X → X/X k ( i ) . As usual, thisinduces a long exact sequence in cohomology · · · → ˜ H ∗ , ∗ (cid:0) X/X k ( i ) (cid:1) → ˜ H ∗ , ∗ ( X ) → ˜ H ∗ , ∗ (cid:0) X k ( i ) (cid:1) → ˜ H ∗ +1 , ∗ (cid:0) X/X k ( i ) (cid:1) → · · · . Observe that, aside from the basepoint, all the cells of the quotient
X/X k ( i ) havefixed-set dimension greater than i . Thus the reduced cohomology of the quotientvanishes for all bidegrees ( p, q ) with p ≤ i and q ≥ p − i − i − pq i Figure 15.
Vanishing region for ˜ H ∗ , ∗ (cid:0) X/X k ( i ) (cid:1) .Fix any bidegree ( p, q ). There exists an i such that i > p and i > p − q −
2. Thisimplies both ( p, q ) and ( p + 1 , q ) lie in the vanishing region for X/X k ( i ) . So in thelong exact sequence for the cofiber sequence X k ( i ) → X → X/X k ( i ) we have · · · → → ˜ H p,q ( X ) → ˜ H p,q (cid:0) X k ( i ) (cid:1) → → · · · and we see that ˜ H p,q ( X ) ∼ = ˜ H p,q (cid:0) X k ( i ) (cid:1) . Similarly, if n > k ( i ) and we consider thecofiber sequence X k ( i ) → X n → X n /X k ( i ) , we have that ˜ H p,q ( X n ) ∼ = ˜ H p,q (cid:0) X k ( i ) (cid:1) .This tells us that, as a vector space, every bidegree of the cohomology of X stabilizesby some finite stage k ( i ). Of course, this value depends on both p and q .Recall, as X = colim X k , we have the Milnor exact sequence0 → lim ←− ˜ H ∗− , ∗ ( X k ) → ˜ H ∗ , ∗ ( X ) → lim ←− ˜ H ∗ , ∗ ( X k ) → , where the inverse limit is taken for the tower˜ H ∗ , ∗ ( X ) ι ∗ ←− ˜ H ∗ , ∗ ( X ) ι ∗ ←− · · · ι ∗ ←− ˜ H ∗ , ∗ ( X k ) ι ∗ ←− ˜ H ∗ , ∗ ( X k +1 ) ι ∗ ←− · · · (see for example Section 3F of [4]). In particular, for each p and q we have theshort exact sequence0 → lim ←− ˜ H p − ,q ( X k ) → ˜ H p,q ( X ) → lim ←− ˜ H p,q ( X k ) → . Fix ( p, q ) and choose i such that i > p and i > p − q − ←− or the inverse limit, so we havelim ←− k ˜ H p − ,q ( X k ) ∼ = lim ←− n ≥ k ( i ) ˜ H p − ,q ( X n )and lim ←− k ˜ H p,q ( X k ) ∼ = lim ←− n ≥ k ( i ) ˜ H p,q ( X n ) . As shown above, the maps in the truncated tower H p,q (cid:0) X k ( i ) (cid:1) ι ∗ ←− H p,q (cid:0) X k ( i )+1 (cid:1) ι ∗ ←− H p,q (cid:0) X k ( i )+2 (cid:1) ι ∗ ←− . . . are all isomorphisms.Recall that lim ←− vanishes whenever all the maps in the tower are surjective andthus lim ←− ˜ H p − ,q ( X k ) = 0. The same argument can be made for any bidegree ( p, q ),hence lim ←− ˜ H ∗− , ∗ ( X k ) ≡
0, making ˜ H ∗ , ∗ ( X ) → lim ←− ˜ H ∗ , ∗ ( X k ) an isomorphism ofbigraded vectors spaces. Since this is an M -module map, ˜ H ∗ , ∗ ( X ) ∼ = lim ←− ˜ H ∗ , ∗ ( X k )as M -modules.It remains to show that lim ←− ˜ H ∗ , ∗ ( X k ) is free. For that we use a stabilizationphenomenon of the cohomologies as M -modules to find a basis for the inverselimit. Fix an i ≥ H ∗ , ∗ (cid:0) X k ( i +1) (cid:1) as an M -module. If we attachthe next cell to form X k ( i +1)+1 and compute its cohomology via the usual cofibersequence, any M generators with fixed-set dimension i will be sent to zero by thedifferential in the long exact sequence. This is simply for degree reasons, as an M generator with fixed-set dimension i cannot support a nonzero differential to an M generated by an element with fixed-set dimension i + 2 or greater. Furthermore,the generators of fixed-set dimension i will continue to support zero differentialswhen computing the cohomology of X n for any n > k ( i + 1).Now for each i ≥
0, choose a graded free basis for ˜ H ∗ , ∗ (cid:0) X k ( i +1) (cid:1) and let β i bea subset of that basis consisting only of generators of fixed-set dimension i . Theunion of all such β i will form our graded free basis for ˜ H ∗ , ∗ ( X ). For any index i , wecan define a map M h β i i → lim ←− ˜ H ∗ , ∗ ( X k ) by including M h β i i → ˜ H ∗ , ∗ (cid:0) X k ( i +1) (cid:1) .In fact, we have an inclusion M h β i i → ˜ H ∗ , ∗ ( X n ) for any n > k ( i + 1) since if agenerator in β i has bidegree ( a, b ), then ˜ H a,b (cid:0) X k ( i +1) (cid:1) ∼ = ˜ H a,b ( X n ). Thus we candefine a cone over the inverse system˜ H ∗ , ∗ (cid:0) X k ( i +1) (cid:1) ˜ H ∗ , ∗ (cid:0) X k ( i +1)+1 (cid:1) ˜ H ∗ , ∗ (cid:0) X k ( i +1)+2 (cid:1) · · · M h β i i ι ∗ ι ∗ ι ∗ and so we get a map to the inverse limit f i : M h β i i → lim ←− j ˜ H ∗ , ∗ (cid:0) X k ( i +1)+ j (cid:1) ∼ = lim ←− ˜ H ∗ , ∗ ( X k ) . Finally taking the direct sum over all f i we have a map f : ∞ M i =0 M h β i i −→ lim ←− ˜ H ∗ , ∗ ( X k ) ∼ = ˜ H ∗ , ∗ ( X ) . REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 23 As above, for any bidegree ( p, q ), the map f factors through a finite stage where˜ H p,q (cid:0) X k ( i ) (cid:1) ∼ = ˜ H p,q ( X ) for some appropriate choice of i . Thus the M -modulemap f is an isomorphism because it is an isomorphism in every bidegree, and hence H ∗ , ∗ ( X ) is free. (cid:3) Corollary 6.4.
Suppose X is a finite type Rep( C ) -complex. Then F (Σ ∞ X + , H F ) ≃ _ i ∈ I (cid:0) S − p i , − q i ∧ H F (cid:1) . Proof.
Let β be a free basis for H ∗ , ∗ ( X ) with each generator γ i having bidegree( p i , q i ) for i ∈ I . By adjunction H ∗ , ∗ ( X ) ∼ = π −∗ , −∗ F (Σ ∞ X + , H F ) via H p,q ( X ) = [Σ ∞ X + , S p,q ∧ H F ] ∼ = [Σ ∞ X + ∧ S − p, − q , H F ] ∼ = [ S − p, − q , F (Σ ∞ X + , H F )]= π − p, − q F (Σ ∞ X + , H F ) . Each γ i ∈ π − p i , − q i F (Σ ∞ X + , H F ) gives rise to a map of spectra S − p i , − q i ∧ H F γ i ∧ id −−−→ F (Σ ∞ X + , H F ) ∧ H F µ −→ H F , where F (Σ ∞ X + , H F ) is an H F -module via the structure map µ . Taking all ofthese maps together, we get a map _ i ∈ I (cid:0) S − p i , − q i ∧ H F (cid:1) −→ F (Σ ∞ X + , H F ) . By construction this map induces an isomorphism on bigraded homotopy. Since RO ( C )-graded homotopy groups detect weak equivalences (see for example Section6.10 of [9]), the map is a weak equivalence. (cid:3) Counterexample 6.5.
The freeness theorem does not hold for all locally finiteRep( C )-complexes. Consider the space Y = W ∞ n =0 S n,n . By the wedge axiom˜ H ∗ , ∗ ( Y ) ∼ = ∞ Y n =0 ˜ H ∗ , ∗ ( S n,n ) ∼ = ∞ Y n =0 Σ n,n M . Nonequivariantly, the singular cohomology of the underlying space is a free F -module because F is a field. However, M is not a field and ˜ H ∗ , ∗ ( Y ) is not a free M -module. Suppose γ n is the generator of the M in bidegree ( n, n ) of the infiniteproduct. Then there is an element in bidegree (0 , −
2) of ˜ H ∗ , ∗ ( Y ) of the form x = (cid:18) θγ , θρ γ , θρ γ , θρ γ , . . . (cid:19) . Notice that x is not ρ -torsion since for any k we have ρ k x = 0. However τ x = 0.In M , and indeed in any free module, an element that is not ρ -torsion is also not τ -torsion. Hence ˜ H ∗ , ∗ ( Y ) cannot be free. Shifting of free generators
In this section we specify the Kronholm shifts of free generators that occur whenattaching a single cell to a Rep( C )-complex with an interesting differential. Byinteresting differential, we mean a differential of the last type that shows up inthe proof of the freeness theorem in Section 6.1, where a ramp of generators mapto the bottom cone. The setup of the following theorem is the same as case (c)in the proof, with X playing the role of X k and Y playing the role of X k +1 . Theresult is that we can precisely specify the bidegrees of free generators in cohomologyresulting from such a differential. Theorem 7.1.
Let X be a finite p -dimensional Rep( C ) -complex. Suppose X isformed by attaching representation cells in order of increasing topological dimensionand increasing weight, where any cells of dimension p have weight less than or equalto q . Let Y be the space formed by attaching a single Rep( C ) -cell D ( R p,q ) to X .The cofiber sequence X → Y → Y /X ∼ = S p,q induces a long exact sequence inreduced cohomology with differential d : ˜ H ∗ , ∗ ( X ) → ˜ H ∗ +1 , ∗ ( S p,q ) . Assume thereexists a basis for ˜ H ∗ , ∗ ( X ) ∼ = M h ω , . . . , ω n , χ n +1 , . . . , χ m +1 i , where ω , . . . , ω n sat-isfy the ramp condition and let ν denote the free generator of ˜ H ∗ , ∗ ( S p,q ) . Supposeeach ω i supports a nonzero differential to M − h ν i with d ( ω i ) = θρ j i τ k i ν for some integers j i , k i ≥ and d ( χ i ) = 0 for n + 1 ≤ i ≤ m + 1 . Then ˜ H ∗ , ∗ ( Y ) ∼ = n M i =1 Σ | ω i | +(0 ,s i ) M ! ⊕ (cid:16) Σ | ν |− (0 ,s + s + ··· + s n ) M (cid:17) ⊕ m +1 M j = n +1 Σ | χ j | M where the shifts in weight are given by s = k + 1 s = k − k s = k − k ... s n = k n − k n − . Proof.
The cofiber sequence X → Y → S p,q induces a long exact sequence incohomology · · · d −→ ˜ H ∗ , ∗ ( S p,q ) π ∗ −→ ˜ H ∗ , ∗ ( Y ) ι ∗ −→ ˜ H ∗ , ∗ ( X ) d −→ ˜ H ∗ +1 , ∗ ( S p,q ) π ∗ −→ · · · as usual.In the proof of Theorem 6.1, we investigated the short exact sequence0 → cok( d ) → ˜ H ∗ , ∗ ( Y ) → ker( d ) → . Recall from part (c) of the proof, using τ -localization we know the topologicaldimensions of the free generators of ˜ H ∗ , ∗ ( Y ) are the same as those of ˜ H ∗ , ∗ ( X )together with ˜ H ∗ , ∗ ( S p,q ). Similarly, we know from ρ -localization that the fixed-setdimensions of the free generators are preserved. Furthermore, we know that the REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 25 pq s s s s s Σ s i ω ω ω ω ω ν pq a a a a a a Figure 16.
Example of Kronholm shifts predicted by Theorem7.1, corresponding to Figures 14, 17, and 18.generator in topological dimension p of ˜ H ∗ , ∗ ( Y ) that “corresponds” to ν has fixed-set dimension at least p − q . We also had the following split short exact sequence0 → M h b n +1 , . . . , b m +1 i → ˜ H ∗ , ∗ ( Y ) → N → , where ι ∗ ( b i ) = χ i , making the quotient N a free direct summand.Putting all of this together we have a short exact sequence of short exact se-quences as below. 0 0 00 0 L i M h b i i L i M h χ i i
00 cok( d ) ˜ H ∗ , ∗ ( Y ) ker( d ) 00 cok( ˜ d ) N ker( ˜ d ) 00 0 0Here ˜ d is the restriction of the differential to the direct summand M h ω , . . . , ω n i .An example of the restricted differential ˜ d was shown in Figure 14. For ease ofreference, the same example appears again on the left in Figure 18.Now fix i and consider an element x ∈ N with bidegree | x | = | ω i | . As a conse-quence of the ramp condition and the fact that each ω i supports a nonzero differ-ential, the kernel of ˜ d in bidegree | ω i | is always zero. So x must be sent to zero inker( ˜ d ) and thus lift to an element of cok( ˜ d ). Because im( d ) ⊆ M − h ν i , any elementof cok( d ) in this bidegree has τ -torsion. We know there is a single free generator of N with topological dimension top( ω i ),but no element of M +2 has τ -torsion. So this free generator of N must have higherweight (and so lower fixed-set dimension). This same argument holds for each ω i .So every generator in these topological dimensions must have “shifted” up. Inorder to preserve fixed-set dimensions, the generator in topological degree p thatcorresponds to ν must have “shifted” down.To clarify with an example, Figure 17 shows a slanted grid corresponding to thetopological and fixed-set dimensions of the generators for the differential picturedin Figure 14 (and again on the left in Figure 18). On the left of Figure 17 we seethe original positions of the generators and on the right the positions of the shiftedgenerators. The red generators correspond to ω , . . . , ω n and the blue generatorcorresponds to ν . A combinatorial argument allows us to identify the positions ofthe shifted generators shown on the right.As there is at most one generator in each topological dimension and each fixed-set dimension, we must have a single generator on each vertical line and a singlegenerator on each diagonal line, both before and after the differential. We have justargued above that all the red generators, those corresponding to ω , . . . , ω n , mustshift up. If we consider the generators from left to right, each will have a uniqueunoccupied diagonal above it on the slanted grid. So finally the blue generatorcorresponding to ν must shift down as pictured in Figure 17. pqq − pqq − Figure 17.
Shifts along the slanted grid.In general, since we have shown each ω i shifts up, again if we consider each topo-logical dimension from left to right, each generator has a unique position availableto it on the slanted grid. These shifts of the ω i and ν are precisely as calculated inthe theorem statement. (cid:3) Identifying shifted generators.
So far we have shown the free generators“shift” in the case of a ramp differential, but the new generators remain a bitmysterious. For the interested reader, we will precisely identify the new shiftedgenerators. To do so we pick out a collection of distinguished elements in the kernelof the differential. These are the lowest-weight elements in M +2 h ω , . . . , ω n i in each REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 27 of the topological degrees top( ω ) , . . . , top( ω n ) , top( ν ) supporting zero differentials.We will use these to define elements in ˜ H ∗ , ∗ ( Y ).The following chart is useful for reference as we define these elements.˜ H ∗ , ∗ ( Y ) ι ∗ −→ ˜ H ∗ , ∗ ( X ) d −→ ˜ H ∗ +1 , ∗ ( S p,q ) a τ k +1 ω a i ρ j i − j i +1 ω i + τ k i +1 − k i ω i +1 a n ρ j n +1 ω n b i χ i a i on the right. pqq − νω ω ω ω ω νω ω ω ω ω a a a a a a Figure 18.
Identifying shifted generators.Notice τ k +1 ω , as well as ρ j i − j i +1 ω i + τ k i +1 − k i ω i +1 for 1 ≤ i ≤ n − ρ j n +1 ω n are all elements of ker( d ). The differential is zero on the first and last ofthese elements for degree reasons, while d ( ρ j i − j i +1 ω i + τ k i +1 − k i ω i +1 ) = 0 becausewe are working mod 2. Since each of these elements is nonzero in ˜ H ∗ , ∗ ( X ), byexactness we can define a , . . . , a n to be their preimages in ˜ H ∗ , ∗ ( Y ) as in the chartabove. These a i are free generators and N ∼ = M h a , . . . , a n i .Each a i is uniquely determined because ι ∗ is injective in bidegree | a i | . Thisfollows because in the long exact sequence · · · → ˜ H | a i |− (1 , ( X ) d −→ ˜ H | a i | ( S p,q ) π ∗ −→ ˜ H | a i | ( Y ) ι ∗ −→ ˜ H | a i | ( X ) → · · · the map d : ˜ H | a i |− (1 , ( X ) → ˜ H | a i | ( S p,q ) is surjective making π ∗ zero in bidegree | a i | and hence ι ∗ injective.As the a i are uniquely determined and we already know N has free generatorsin the same bidegrees as the a i , each a i is indeed a free generator. The otherdistinguished elements of the kernel that we can lift back to ˜ H ∗ , ∗ ( Y ) are the χ i . This is clear from Figure 18 and follows from the image under d of ρ j i − j i +1 − τω i for any1 ≤ i ≤ n − ρ j n τω n since the rank of ˜ H ∗ , ∗ ( S p,q ) is at most one in any bidegree. Since ˜ H ∗ , ∗ ( Y ) ∼ = N ⊕ M h b n +1 , . . . , b m +1 i , the elements a , . . . , a n , b n +1 , . . . , b m +1 give a free basis for ˜ H ∗ , ∗ ( Y ).8. Kronholm’s proof
In this section, we discuss the subtle gap in Kronholm’s proof of the freenesstheorem and compare his approach with the proof in Section 6. Kronholm’s proof in[6] also proceeds by induction on the representation cells of a Rep( C )-complex. Asin our proof, the representation cells are attached in order of increasing topologicaldimension and increasing weight. The proof presented in this paper uses the longexact sequence for attaching a single representation cell. Kronholm’s proof makesuse of the following spectral sequence for each weight q . Proposition 8.1.
Let X be a filtered C -space. Then for each q ∈ Z there is aspectral sequence with E p,n = ˜ H p,q (cid:0) X n +1 , X n (cid:1) converging to ˜ H p,q ( X ) . It is convenient to plot the spectral sequences for all q at once in the planeand refer to them collectively as “a spectral sequence” that converges to ˜ H ∗ , ∗ ( X ).Again we plot topological dimension along the horizontal axis and weight alongthe vertical axis. The differential on each page of the spectral sequence increasestopological degree by one as before, but also reaches farther up in the filtrationon each page. In order to depict the spectral sequence in the plane, we suppressthe filtration degree. It is often helpful to use different colors to keep track of thefiltration degree of particular groups.Notice that a cofiber sequence of the form X ֒ → Y → Y /X is a filtration of Y with only two parts X ⊆ Y , giving a degenerate case of the spectral sequencethat converges almost immediately with E ∼ = E ∞ . This degenerate spectral se-quence corresponds closely to the long exact sequence associated to the cofibersequence. In the degenerate spectral sequence, on the E page we have the dif-ferential ˜ H ∗ , ∗ ( X ) d −→ ˜ H ∗ +1 , ∗ ( Y /X ), which is the connecting homomorphism in thelong exact sequence. After taking cohomology with respect to this differential, the E ∼ = E ∞ page is an associated graded module of ˜ H ∗ , ∗ ( Y ). That is, to compute˜ H ∗ , ∗ ( Y ) requires solving the usual extension problem0 → cok( d ) → ˜ H ∗ , ∗ ( Y ) → ker( d ) → . Starting with the ‘one-at-a-time’ cellular filtration X ⊆ · · · ⊆ X k ⊆ · · · ⊆ X where each X k is formed from X k +1 by attaching a single cell, there are twoapproaches one could use to prove the freeness theorem:(A) Use the spectral sequence for the filtration X ⊆ · · · ⊆ X k ⊆ · · · ⊆ X ,analyze the differentials on each page, and solve the extension problems onthe E ∞ page to prove the freeness of H ∗ , ∗ ( X ).(B) Inductively assume H ∗ , ∗ ( X k ) is free, use the long exact sequence associatedto the cofiber sequence X k ֒ → X k +1 → S p k +1 ,q k +1 (or equivalently thedegenerate spectral sequence for the filtration X k ⊆ X k +1 ), and solve theusual extension problem to show the freeness of H ∗ , ∗ ( X k +1 ). See Proposition 3.1 in [6].
REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 29 The pictures one would draw in each of the two approaches are very similar, withsome subtle differences. The main problem with Kronholm’s proof is conflatingthese two approaches. His proof applies reasoning that would be correct for (A) inan incorrect way to (B).Kronholm’s proof begins with approach (B). By induction one may assume that˜ H ∗ , ∗ ( X k ) is free and proceed to show that ˜ H ∗ , ∗ ( X k +1 ) is free. His proof goeson to use a change of basis lemma (much like Lemma 5.3) to change the basisfor ˜ H ∗ , ∗ ( X k ). This change of basis must occur in setting (B), since at no pointin approach (A) does ˜ H ∗ , ∗ ( X k ) appear. After his change of basis, the nonzerodifferential to M h ν i ∼ = ˜ H ∗ , ∗ ( X k +1 /X k ) in the long exact sequence is supported bya ramp of basis elements ω , . . . , ω n .The issue is that Kronholm then further assumes one may consider a nonzerodifferential supported by only a single M generator ω , which is not always possiblein approach (B). The apparent justification for the reduction comes from approach(A). That is, on the E page of the spectral sequence corresponding to the ‘one-at-a-time’ cellular filtration for X k +1 , the only possible differential to M h ν i comesfrom one filtration degree lower, where there is at most a single copy of M from˜ H ∗ , ∗ ( X k /X k − ). Notice this is a different setting from the one where the originalchange of basis occurred. In addition, in approach (A) there may be other differ-entials on the E page, as long as they only increase filtration degree by one. Theconsequences of these possible differentials have not been addressed. See Figure19 for examples of E pages and possible differentials for the two different spectralsequences. pqq − νγ γ γ E for X ⊆ X ⊆ X ⊆ X ⊆ X pqq − νω ω ω ω E for X k ⊆ X k +1 Figure 19.
Examples of spectral sequences for the two approaches.Even if the differentials on the E page in approach (A) can be resolved, it isnot clear how to induct because the E page and other subsequent terms of thespectral sequence will usually not be free M -modules. Examples of such non-free“Jack-O-Lantern” modules appear in [5]. In particular, the E ∞ page is generally not free. Rather, as a direct consequence of the main theorem, Theorem 6.1, the E ∞ page is an associated graded module of a filtration for a free M -module.There are multiple approaches one might use to correct the proof. In this paperwe have chosen approach (B). We consider the long exact sequence associated to thecofiber sequence for attaching a single cell X k ֒ → X k +1 → S p k +1 ,q k +1 and assumeby induction that ˜ H ∗ , ∗ ( X k ) is free. We show that when a ramp of basis elements ω , . . . , ω n support nonzero differentials, the resulting cohomology ˜ H ∗ , ∗ ( X k +1 ) isstill free. We use the second author’s structure theorem to simplify the argument.A proof with approach (B) using similar techniques to the one written by Kronholmwould also work, but would require a lot of bookkeeping. Alternatively, one couldtake approach (A) and investigate the consequences of various differentials on eachpage of the ‘one-at-a-time’ cellular spectral sequence.Yet another approach might be to show that every Rep( C )-complex X has acellular filtration for which E ∼ = ˜ H ∗ , ∗ ( X ) so that the cellular spectral sequence forthis filtration has only zero differentials. This last approach is motivated by thefact that a number of computations are simplified by finding a cellular filtrationfor which all differentials are zero. A possible counterexample is the equivariantGrassmannian Gr ( R , ). There are a number of cellular filtrations (and so various E pages) for Gr ( R , ) using the usual Schubert cell decomposition and differentflag symbols for R , . For example, we can decompose R , ∼ = R ++ −− ∼ = R + − + − (see [6] for details).By comparing the equivariant Schubert cell decompositions and possible differ-entials, one can compute ˜ H ∗ , ∗ (Gr ( R , )) as an M -module. However, the resultingcohomology is not isomorphic to any of the E pages coming from such a Schubertcell decomposition. Each of these E pages has at least one nonzero differen-tial. If it is true that every Rep( C )-complex X has a cellular filtration for which E ∼ = ˜ H ∗ , ∗ ( X ), there must be yet another Rep( C )-complex structure for Gr ( R , ).To illustrate approach (A) and some of the challenges it may present, as wellas the source of the confusion, we return to the computation in Example 3.2,˜ H ∗ , ∗ ( R P tw ). Example 8.2. As R P tw = P ( R , ) = Gr ( R , ) is an example of an equivariantGrassmannian, we consider the Schubert cell decompositions for R P tw as Gr ( R ++ − )and as Gr ( R + − + ). The first will give a cell decomposition with cells in bidegrees(0 , , (1 ,
0) and (2 , , , (1 ,
1) and(2 , E pages for each decomposition areshown in Figure 20. In these pictures we use color to keep track of filtration degree.The possible differentials on these E pages are also depicted in Figure 20. Wewill see the dashed d is zero by comparing the E pages on the left and right since˜ H ∗ , ∗ (Gr ( R ++ − )) ∼ = ˜ H ∗ , ∗ (Gr ( R + − + )). The E page on the right has a zero inbidegree (2 ,
0) and no differential could change that. So on the left there must bea nonzero differential from the generator in bidegree (1 ,
0) to θ times the generatorin bidegree (2 , E page on the right, if the resulting cohomology is toagree with the left, the differential on the generator in (1 ,
1) must be zero, or theresulting isomorphism with the M in bidegegree (2 ,
1) would yield a contradiction.Taking cohomology we have the following E pages in Figure 21.In both cases, d must be zero, since it would increase filtration degree by two.On the right, the E ∞ page is free so the extension problem is easily solved, and we REENESS THEOREM FOR COHOMOLOGY OF REP( C )-COMPLEXES 31 pq − − − − − − − d E for Gr ( R ++ − ) pq − − − − − − − d = 0 E for Gr ( R + − + ) Figure 20.
Two E pages for R P tw . pq − − − − − − − E ∼ = E ∞ for Gr ( R ++ − ) pq − − − − − − − E ∼ = E ∞ for Gr ( R + − + ) Figure 21.
Two E ∼ = E ∞ pages for R P tw .have E ∼ = E ∞ ∼ = ˜ H ∗ , ∗ ( R P tw ) ∼ = Σ , M ⊕ Σ , M as in Example 3.2. On the left, E ∼ = E ∞ is an associated graded M -module fora filtration of ˜ H ∗ , ∗ ( R P tw ), but we see E ∞ = ˜ H ∗ , ∗ ( R P tw ). In fact, looking at the E ∞ page on the left, it is rather surprising that the cohomology of R P tw is a free M -module.Notice this computation agrees with the long exact sequence and cohomologyfrom Example 3.2. It is precisely this similarity between the long exact sequenceand the cellular spectral sequence that is the source of the trouble. If we had athird cell in a higher filtration degree, the differential on the E page could be more complicated. In addition, we would need to consider possible differentials on the E page, which may not be so easily determined, especially if the E page is notfree. References [1] D. Dugger. Bigraded cohomology of Z / Geom. Topol. ,19(1):113–170, 2015.[2] K. K. Ferland.
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